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  "body": "\n## Proof\n\nWe multiply the two series by the Cauchy product and use the binomial theorem.\n\n1. Let $x, y$ be real. Let $a_k = \\frac{x^k}{k!}$ and $b_k = \\frac{y^k}{k!}$ for $k \\ge 0$. By the definition of the [[exponential function|0RMVS7RXZX]], $\\sum a_k = \\exp(x)$ and $\\sum b_k = \\exp(y)$.\n2. Both series converge absolutely, since the [[exponential series has infinite radius of convergence|0PVJRACB9H]].\n3. The [[Cauchy product|0B1K9PVWE9]] of the two series has the terms $c_n = \\sum_{k=0}^{n} a_k b_{n-k} = \\sum_{k=0}^{n} \\frac{x^k y^{n-k}}{k! \\, (n-k)!}$.\n4. By the [[factorial formula for binomial coefficients|033KTKKRV6]], $\\frac{1}{k! \\, (n-k)!} = \\frac{1}{n!} \\binom{n}{k}$ for $0 \\le k \\le n$.\n5. By step 4 and the [[binomial theorem|0EHWPWGHB2]] in $\\R$, $c_n = \\frac{1}{n!} \\sum_{k=0}^{n} \\binom{n}{k} x^k y^{n-k} = \\frac{(x + y)^n}{n!}$.\n6. By step 2 and the theorem on the [[Cauchy product of absolutely convergent series|0AN28XQ45Z]], $\\sum c_n$ converges, and its sum is $\\exp(x) \\exp(y)$.\n7. By step 5 and the definition of $\\exp$, $\\sum_{n=0}^{\\infty} c_n = \\sum_{n=0}^{\\infty} \\frac{(x+y)^n}{n!} = \\exp(x + y)$.\n8. Comparing steps 6 and 7 gives $\\exp(x + y) = \\exp(x) \\exp(y)$.\n\n## Idea\n\nMultiplying the two series and collecting the terms of total degree $n$ gives exactly the binomial expansion of $(x + y)^n$, divided by $n!$. Absolute convergence makes this rearrangement legitimate.\n",
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